NOTE: This answer is extracted from an old key. Sorry for the reference to “problem #24”—just
ignore than and ask me if you need help. I also solve this problem in the key for “initial rate of
reaction” rather than for “initial rate of formation” (and I reworded the problem on the handout
to read). Hopefully this is helpful nonetheless.
44. Data corresponding to the chemical reaction described by C2H4(g) + O3(g)  2 CH2O +
[O3]o (M)
0.50 x 10
-7
[C2H4]o (M)
1.0 x 10
-1
Initial Rate of formation of CH2O (Ms )
-8
1.0 x 10
-12
1
2
O2:
Initial Rate of Reaction
0.50 x 10-12
1.5 x 10-7
1.0 x 10-8
3.0 x 10-12
1.5 x 10-12
1.0 x 10-7
2.0 x 10-8
4.0 x 10-12
2.0 x 10-12
DETAILED SOLUTION AND COMMENTARY.
(a) To determine the rate law basically* means to “determine the exponents of the concentration terms in
m
n
the general rate law: Rate = k[A] [B] ….” which in this problem means finding “m” and “n” in:
m
n
*I’d say that it also means finding the value of
Rate = k[O3] [C2H4]
k, but that’s not always the interpretation.
The easiest way to start this kind of problem is to realize that m corresponds to “how the rate
of reaction depends on the concentration of O3” and that n corresponds to “how the rate of
reaction depends on the concentration of C2H4”. That means look for two trials in which only
one of the concentration values has changed. Then you can compare how the rate changed
with how the (single) concentration value changed.
As in my answer to #24, I will give the “verbal” (calculations in your head) way to look at this
problem first, and then I will give the “brute force” algebra approach. I hope that with practice
you will become familiar “seeing” how the math works so that you don’t need to do the brute
force algebra approach. On the other hand, I want you to feel completely comfortable using
the algebraic method as well! 
A. Verbal analysis. In this case, first look at trials 1 and 2 since [C2H4] does not change.
The concentration of O3 tripled and the (initial) rate (of formation) ALSO TRIPLED. Right
away that should tell you that m must be 1—the rate of formation is directly proportional to
the concentration of O3 (to the first power). (See the answer to #24 for more details on why
this is so.) NOTE THAT THIS IS JUST THE OPPOSITE OF PROBLEMS 24 AND 26! It will
be good for you to become comfortable being able to “see” these kinds of problems from
“both directions”. Please ask for help if necessary!
Okay, finding n is not as simple in this problem! Why? You hopefully noticed that there is
no pair of trials in which only the concentration of C2H4 changed!!!! THIS DOES NOT MEAN
THAT YOU CANNOT DETERMINE n! Why? Because you already KNOW the order with
respect to O3! Hopefully you will see how this removes the “restriction” that you have two
trials with the same concentration of O3. You must compare trials 1 and 3 or trials 2 and 3 to
get information about n since those are pairs of trials in which the concentration of C2H4
changed. One way to reason this out (other than the “brute force” algebraic method noted
below) is to first FIGURE OUT THE FACTOR OF CHANGE IN RATE THAT SHOULD HAVE
OCCURRED AS A RESULT OF THE CHANGE IN CONCENTRATION OF [O3] since in these
pairs of trials the O3 concentration does not remain constant.
Let’s work with trials 1 and 3 since the rate ratio there is a whole number (4:1) rather than a
fraction (3:4 or 4:3). In going from trial 1 to trial 3, the concentration of O3 doubled. From our
analysis above, we know that the rate should double as a result of the change in O3
concentration, because we determined that the exponent m is just “one”. But we notice that
the rate did NOT double, it became FOUR TIMES as great. That is, there is an “extra factor
of two” present, so that must be a result of the change in concentration of C2H4. We are sort
of calculating what the factor of change would have been if the concentration of O3 had not
changed! That is, it would have been a factor of 2. And by what factor
did the concentration
of C2H4 change? A factor of two as well (it doubled, from 1.0 x 10-8 M to 2.0 x 10-8 M). So
that means that the exponent of [C2H4] must also be a “one”.
Is this result consistent with the changes from trial 2 to trial 3? Yes! The concentration of O3
became 2/3 of what it was (1.0 is 2/3 of 1.5) and the concentration of C2H4 became 2 times
what it was, so if both of the exponents are “one”, the rate should just become 2/3 x 2 = 4/3
times the original rate, which it does! (Again, see answer to #24 to follow this “factor” logic.)
Another way to reason this out without doing the “brute force” calculation below was
discussed with a few students in lab on Wednesday and also in the email I sent out on this
problem last week. I will not discuss that method here.
B. Brute force approach. The basic idea is to use the general rate law relation to generate
a pair of equations corresponding to a pair of trials and then use algebra (divide one whole
equation by the other) to solve for the unknown exponent (if you pick two trials in which only
one concentration value changes, then you will only have one “unknown” exponent to solve
for!).
To solve for “m” in this problem. Pick trials 1 and 2 as above ([C2H4] doesn’t change) and
substitute values into the general rate equation: rate = k[O3]m[C2H4]n:
Trial 1: 1.0 x 10-12 Ms-1 = k(0.50 x 10-7 M)m(1.0 x 10-8 M)n
Trial 2: 3.0 x 10-12 Ms-1 = k(1.5 x 10-7 M)m(1.0 x 10-8 M)n
There are multiple ways to use algebra to solve this “system of two equations”. The most
straightforward way is to divide the equation having the larger rate in it by the one with
the smaller rate (I find it easier to work with values greater than one rather than
fractions!). (Note that this works simply because if a = b and c = d, then it follows that a/c
= b/d.). So, dividing Trial 2’s equation by Trial 1’s, you get (after canceling out a bunch
of identical terms and units in the numerator and denominator of both sides):
3.0 (1.5 x10-7 M)m

1.0 (0.50 x10-7 M)m
a
xa  x 
Recall that (xy) = x y , and a    so the above simplifies to:
y
y
a
a a
m
3.0 (1.5)m (10-7 M)m
 1.5 
m

 3.0  
  3.0  3.0
m
-7
m
1.0 (0.50) (10 M)
 0.50 
from which it is clear that m must be 1.
To solve for “n” in this problem. Pick Trials 1 and 3 (or 2 and 3) and substitute into the rate
equation as above except that now you know that m = 1! You see, once you know the
value of m, you don’t have to worry about whether or not the concentration of O3
changes!
Trial 1: 1.0 x 10-12 Ms-1 = k(0.50 x 10-7 M)1(1.0 x 10-8 M)n
Trial 3: 4.0 x 10-12 Ms-1 = k(1.0 x 10-7 M)1(2.0 x 10-8 M)n
Dividing Trial 3’s equation by Trial 1’s, you get (recalling that
ab a b
 x ):
cd c d
4.0 (1.0 x10-7 M)1 (2.0 x10-8 M)n

x
1.0 (0.50 x10-7 M)1 (1.0 x10-8 M)n
 4.0 
Rate became 4 times
what it was (in going from
Trial 1 to Trial 4).
2
x
2n
Factor of 2 came from
the change in
concentration of O3
Factor of 2n came from the
change in concentration of
C2H4
n
So clearly, n = 1. (2 = 2 ) and thus the rate law is: rate = k[O3][C2H4]
(**NOTE: In lab (and possibly on an exam), you will need to know how to solve for n if it is not
“obvious” or an integer value (due to experimental error or uncertainty). For example, to solve for n
in
n
2.13 = 2.00 ,
you need to “take the log (or ln) of both sides” to obtain:
log(2.13) = log(2.00)n
which, using the property of logs that log(x)n = n(log(x)) becomes
log(2.13) = n(log(2.00))
so that you can solve for n:
n=
log(2.13) 0.328

 1.09  1.1 )
log(2.00) 0.301
(b) The term “order” refers to the value of the exponent. Since the exponents here are “one” for both
reactants, the order is 1 for both reactants (or “first order” with respect to O3 and C2H4)
(c) Solving for k in the rate law expression gives:
-12
-1
k=
If you used “rate of formation of CH2O” (1.0 x 10 Ms ) rather than the
3
-1
-1
“rate of reaction” here, you would get 2.0 x 10 M s for k.
Using trial 1: k =
initial rate of reaction (for a given trial)
[O 3 ][C 2H4 ] (for that trial)
0.50 x 10-12 M  s-1
 1000 M-1  s-1  1.0 x 103 M-1  s-1
(0.50 x 10-7 M)(1.0 x 10-8 M)
**Normally you would calculate the k value for all trials and average them since they may
not be identical to the number of digits that are justified. Here all values turned out to be
the same so I will not show the calculations. Note that the units of k are not always
the same; you need to figure those out by analyzing the units in each calculation of
a “k”. Note also that you should be careful to indicate precision properly (write the
number of digits that indicates properly where the uncertainty lies.**
(d) Once the rate law and “k” are known, you can calculate (i.e., “predict”) what the rate of reaction
ought to be for a given set of concentrations (as long as the T is kept the same, of course)! Just
substitute the concentration values into the rate law!
Rate (of rxn) = (1.0 x 103 M-1s-1)[O3][C2H4] = (1.0 x 103 M-1s-1)(2.0 x 10-7 M)(2.0 x 10-7 M) =
4.0 x 10-11 Ms-1
Reminder (again): The rate just calculated is actually the rate of reaction (asked for in the
problem), which is not the same as the “rate of formation of CH2O” (which is what many of you may
have calculated). The rate of formation of CH2O is twice the rate of reaction in this case because
the coefficient of CH2O is 2 in the balanced equation. So many of you probably ended up with 8.0
-1
x 10-11 Ms .